有什么理由分配-0？
Any reason to assign -0?
我正在使用 Lint 检查一些旧的 C 代码，偶然发现了这一行：
int16_t max = -0;
Lint 消息是 "Constant expression evaluates to 0 in operation '-'"。
有人会使用 -0 有什么原因吗？
在 C 规范 （6.2.6.2 整数类型） 中，它声明如下（强调我的）：
For signed integer types, the bits of the object representation shall be divided into three groups: value bits, padding bits, and the sign bit. There need not be any padding bits; there shall be exactly one sign bit. Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type (if there are
M value bits in the signed type and N in the unsigned type, then M £ N). If the sign bit is zero, it shall not affect the resulting value. If the sign bit is one, the value shall be modified in one of the following ways:
- the corresponding value with sign bit 0 is negated (sign and
magnitude);
- the sign bit has the value -(2N) (two’s complement);
- the sign bit has the value -(2N - 1) (one’s complement).
Which of these applies is implementation-defined, as is whether the
value with sign bit 1 and all value bits zero (for the first two), or
with sign bit and all value bits 1 (for one’s complement), is a trap
representation or a normal value. In the case of sign and magnitude
and one’s complement, if this representation is a normal value it is
called a negative zero.
也就是说，C支持三种不同的representations for signed integers and two of them have the concept of signed zero，区分了正零和负零
所以，我的解释是，您的代码片段的作者可能试图生成负零值。但是，正如中所指出的，这个表达式实际上不能产生负零，这意味着作者可能在那里做出了错误的假设。
不，我看不出有任何理由。其他人提到可以有 "negative zero" 的平台，但是这个表达式永远不会产生这样的负零，所以这是没有用的。
C标准中对应的段落是6.2.6.2 p3，重点是我的：
If the implementation supports negative zeros, they shall be generated
only by:
— the &, |, ^, ~, <<, and >> operators with operands that
produce such a value;
— the +, -, *, /, and % operators where one operand is a negative zero and the result is zero;
— compound
assignment operators based on the above cases.
例如，要在这样的平台上生成负零，您可以使用 ~INT_MAX，但对于其他表示，这不会是零，因此代码不是很便携。
这是针对使用 CPU 和 one's complement 数字的架构。
- 补码是一种表示负数的方式，具有 -0。仍在使用浮点数。
- 对于无符号数，最大数确实是
-0：全 1。
我们习惯了补码，多了一个负数。虽然一个人的补码在零附近有一些麻烦，但同样适用于 MIN_INT 附近的二进制补码：-MIN_INT == MIN_INT.
上面的代码可能是无符号数：
uint16_t max = (uint16_t) -1;
C99 或更高版本没有理由。
int16_t 是精确宽度的整数类型。
The typedef name intN_t designates a signed integer type with width N, no padding bits, and a two’s complement representation. Thus, int8_t denotes such a signed integer type with a width of exactly 8 bits. C11dr §7.20.1.1 1
二进制补码中没有带符号的零。所以代码相当于
int16_t max = 0;
IMO，int16_t max = -0; 希望 的结果，在非 2 的补码平台上，是将 max 初始化为 -0 到标记长度为 0 或仅包含值为 -0.
的元素的数组
我正在使用 Lint 检查一些旧的 C 代码，偶然发现了这一行：
int16_t max = -0;
Lint 消息是 "Constant expression evaluates to 0 in operation '-'"。
有人会使用 -0 有什么原因吗？
在 C 规范 （6.2.6.2 整数类型） 中，它声明如下（强调我的）：
For signed integer types, the bits of the object representation shall be divided into three groups: value bits, padding bits, and the sign bit. There need not be any padding bits; there shall be exactly one sign bit. Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type (if there are M value bits in the signed type and N in the unsigned type, then M £ N). If the sign bit is zero, it shall not affect the resulting value. If the sign bit is one, the value shall be modified in one of the following ways:
- the corresponding value with sign bit 0 is negated (sign and magnitude);
- the sign bit has the value -(2N) (two’s complement);
- the sign bit has the value -(2N - 1) (one’s complement).
Which of these applies is implementation-defined, as is whether the value with sign bit 1 and all value bits zero (for the first two), or with sign bit and all value bits 1 (for one’s complement), is a trap representation or a normal value. In the case of sign and magnitude and one’s complement, if this representation is a normal value it is called a negative zero.
也就是说，C支持三种不同的representations for signed integers and two of them have the concept of signed zero，区分了正零和负零
所以，我的解释是，您的代码片段的作者可能试图生成负零值。但是，正如
不，我看不出有任何理由。其他人提到可以有 "negative zero" 的平台，但是这个表达式永远不会产生这样的负零，所以这是没有用的。
C标准中对应的段落是6.2.6.2 p3，重点是我的：
If the implementation supports negative zeros, they shall be generated only by:
— the &, |, ^, ~, <<, and >> operators with operands that produce such a value;
— the +, -, *, /, and % operators where one operand is a negative zero and the result is zero;
— compound assignment operators based on the above cases.
例如，要在这样的平台上生成负零，您可以使用 ~INT_MAX，但对于其他表示，这不会是零，因此代码不是很便携。
这是针对使用 CPU 和 one's complement 数字的架构。
- 补码是一种表示负数的方式，具有 -0。仍在使用浮点数。
- 对于无符号数，最大数确实是
-0：全 1。
我们习惯了补码，多了一个负数。虽然一个人的补码在零附近有一些麻烦，但同样适用于 MIN_INT 附近的二进制补码：-MIN_INT == MIN_INT.
上面的代码可能是无符号数：
uint16_t max = (uint16_t) -1;
C99 或更高版本没有理由。
int16_t 是精确宽度的整数类型。
The typedef name
intN_tdesignates a signed integer type with width N, no padding bits, and a two’s complement representation. Thus,int8_tdenotes such a signed integer type with a width of exactly 8 bits. C11dr §7.20.1.1 1
二进制补码中没有带符号的零。所以代码相当于
int16_t max = 0;
IMO，int16_t max = -0; 希望 的结果，在非 2 的补码平台上，是将 max 初始化为 -0 到标记长度为 0 或仅包含值为 -0.